Describe what is seen when excess ammonia solution is added gradually to copper(II)chloride solution. Give the formula of the complex ion formed.
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A copper(II) salt like CuCl2 will dissolve in water and form a complex hexaaquacopper ion as shown below:
CuCl2 + 6H2O --> [Cu(H2O)6]2+ + 2Cl-
The hexaaquacopper ion is a complex ion meaning that it is an ion with a metal atom in the center (in this case copper) surrounded by several coordinating ligands (in this case 6 water molecules). The entire complex has a 2+ charge and is highly soluble in water, thus giving rise to the light blue color associated with copper salts in water. When ammonia (NH3) is added gradually, the initial reaction that occurs is an acid/base reaction where the ammonia acts as a base and converts 2 of the water molecules to hydroxide ions. This is shown in the equation below:
[Cu(H2O)6]2+ + 2NH3 --> [Cu(H2O)4(OH)2] + 2(NH4)+
As you can see, 2 of the water molecules have been deprotonated by the ammonia to form 2 hydroxide (OH) ions. This [Cu(H2O)4(OH)2] is a neutral complex ion meaning it has no charge. It will precipitate from the solution as a light blue solid.
As excess ammonia is added to the solution, a new reaction will occur where the ammonia acts as a ligand and will replace water in the complex ion as shown below:
[Cu(H2O)6]2+ + 4NH3 --> [Cu(NH3)4(H2O)2]2+ + 4H2O
This tetraammoniadiaquacopper complex ion is highly soluble in water and also a very deep blue color. So as more ammonia is added to the solution the inital light blue precipitate [Cu(H2O)4(OH)2] will dissolve as [Cu(NH3)4(H2O)2]2+ becomes the only complex ion present and the solution turns a very deep blue.
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